Lecture 34, Page 1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Physics 2211: Lecture 34 l...

Lecture 34, Physics 2211 Spring 2005© 2005 Dr. Bill Holm

Physics 2211: Lecture 34Physics 2211: Lecture 34

Rotational KinematicsAnalogy with one-dimensional kinematics

Kinetic energy of a rotating system Moment of inertia


RotationRotation

Up until now we have gracefully avoided dealing with the rotation of objects. We have studied objects that slide, not roll.We have assumed pulleys are without mass.

Rotation is extremely important, however, and we need to understand it!

Most of the equations we will develop are simply rotational analogs of ones we have already learned when studying linear kinematics and dynamics.


ExampleExampleRotationsRotations

Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.Klyde’s angular velocity is:

(1)(1) the same as Bonnie’s

(2)(2) twice Bonnie’s

(3)(3) half Bonnie’s


ExampleExample RotationsRotations

The angular velocity of any point on a solid object rotating about a fixed axis is the same.Both Bonnie & Klyde go around once (2 radians) every two seconds.

(Their “linear” speed v will be different since v = r).

BonnieKlyde V21

V


Rotational VariablesRotational Variables

Rotation about a fixed axisConsider a disk rotating about

an axis through its center

dtd

First, recall what we learned aboutUniform Circular Motion:

dtdx

v (Analogous to )

angular velocity, rad/s



Now suppose can change as a function of time. We define the

angular acceleration:2

2

dt

d

dt

d

Consider the case when is constant. We can integrate this to find and as a function of time:

t

0

constant

200 2

1tt



Recall also that for a point at a distance R away from the axis of rotation:s = R ( in radians)v = R

And taking the derivative of this we find:a = R

t

0

constant

200 2

1tt

R

v

s


Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

And for a point at a distance R from the rotation axis:

s = Rv = Ra = R

2 20 02

210 0 2s s v t at

2 20 02v v a s s

constantconstanta

AngularLinear

0v v at

210 0 2t t

0 t


Example: Wheel And RopeExample: Wheel And Rope A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such

that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

aa

R


Wheel And RopeWheel And Rope Use a = R to find :

= a / R = 4 m/s2 / 0.4 m = 10 rad/s2

Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester.

= 0 + 0(10) + (10)(10)2 = 500 rad212

0 t21

t 0

aa

R

1500

280

rot rad x

radrev


Physics 2211: Lecture 35Physics 2211: Lecture 35

Moments of Inertia Discrete particlesContinuous solid objects

Parallel axis theorem


Rotation & Kinetic EnergyRotation & Kinetic Energy Consider the simple rotating system shown below.

(Assume the masses are attached to the rotation axis by massless rigid rods).

The kinetic energy of this system will be the sum of the kinetic energy of each piece:

vv4

vv1

vv3

vv2

K m vi ii

1

22 So

But vi = ri

or K 12

2I Moment of InertiaMoment of Inertia

(rotational inertial)

about the rotation axis

I m ri ii

2

( I has units of kg m2.)

2 2 21 1

2 2

i i i i

i i

K m r m r

rr1

rr2rr3

rr4

m4

m1

m2

m3


Rotation & Kinetic EnergyRotation & Kinetic Energy

The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating System Rotating System

v is “linear” velocity

m is the mass.

is angular velocity

I is the moment of inertia

about the rotation axis.

21

2K I21

2K mv

2i i

i

I m r


So where

Moment of InertiaMoment of Inertia(Rotational Inertia)(Rotational Inertia)

Notice that the moment of inertia I depends on the distribution of mass in the system.The further the mass is from the rotation axis, the bigger

the moment of inertia.

For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass).

We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics!

21

2K I 2

i ii

I m r


Calculating Moment of InertiaCalculating Moment of Inertia

We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is:

Example: Calculate the moment of inertia of four point masses

(m) on the corners of a square whose sides have length L,

about a perpendicular axis through the center of the square:

mm

mm

L

where ri is the distance from the mass

to the axis of rotation.

2

1

N

i ii

I m r



The squared distance from each point mass to the axis is:

mm

mm

Lr

L/2

2L

m42L

m2L

m2L

m2L

mrmI22222N

1i

2ii

so

I = 2mL2

2L

2L

2r22

2

Using the Pythagorean Theorem



Now calculate I for the same object about an axis through the center, parallel to the plane (as shown):

mm

mm

L

r

4L

m44L

m4L

m4L

m4L

mrmI22222N

1i

2ii

I = mL2



Finally, calculate I for the same object about an axis along one side (as shown):

mm

mm

L

r

2222N

1i

2ii 0m0mmLmLrmI

I = 2mL2



For a single object, I clearly depends on the rotation axis!!

L

I = 2mL2I = mL2

mm

mm

I = 2mL2



For a discrete collection of point masses we found:

I m ri ii

N2

1

r

dm

I r dm2

For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.

We have to do anintegral to find I :


Moments of InertiaMoments of Inertia

Some examples of I for solid objects:

Thin hoop (or cylinder) of mass M and

radius R, about an axis through its center,

perpendicular to the plane of the hoop.

I MR 2

R

I 1

22MR

Thin hoop of mass M and radius R,

about an axis through a diameter.

R


Solid sphere of mass M and radius R,

about an axis through its center.

I 2

52MR

R

I 1

22MR

R

Solid disk or cylinder of mass M and

radius R, about a perpendicular axis

through its center.

Moments of InertiaMoments of Inertia

Some examples of I for solid objects:


ExampleExample Moment of InertiaMoment of Inertia

Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold.Which one has the biggest moment of inertia about an axis through its center?

same mass & radius

solid hollow

(1) solid aluminum (2) hollow gold (3) same


ExampleExample Moment of InertiaMoment of Inertia

Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center.The spherical shell (gold) will have a bigger moment of inertia.

ISOLID < ISHELL

same mass & radius

solid hollow


More Moments of InertiaMore Moments of Inertia

Some examples of I for solid objects

Thin rod of mass M and length L, about

a perpendicular axis through its center.

I 1

122ML

L


a perpendicular axis through its end.

I 1

32ML

L


Moment of InertiaMoment of InertiaCalculation ExampleCalculation Example


a perpendicular axis through its end.

I 1

32ML

L

x

y

dx

dm dx is linear mass density M L

2

0

L

I x dm 2

0

L

x dx 2

0

LMx dx

L

3

03

LM x

IL

21

3ML


Parallel Axis TheoremParallel Axis Theorem

Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, ICM, is known.

The moment of inertia about an axis parallel to this axis but a distance D away is given by:

IPARALLEL = ICM + MD2

So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.


Parallel Axis Theorem: ExampleParallel Axis Theorem: Example

Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod.

ICM ML1

122

We know

IEND ML ML

ML

1

12 2

1

32

22So

which agrees with the result on a previous slide.

L

D=L/2M

xCM

ICMIEND

IPARALLEL = ICM + MD2

Lecture 34, Page 1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Physics 2211: Lecture 34 l...

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